Problem: What is the value of the following logarithm? $\log_{6} \left(\dfrac{1}{36}\right)$
Answer: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $6^{y} = \dfrac{1}{36}$ In this case, $6^{-2} = \dfrac{1}{36}$, so $\log_{6} \left(\dfrac{1}{36}\right) = -2$.